Question

A flywheel rotating at an initial velocity of 2.09 rad/s undergoes a change to 4.6 rad/s for precisely 5s

a) Calculate the angular acceleration of the flywheel.

b) Suppose the flywheel has moment of inertia of 1.5 kg m^2. Compute the wheels torque?
[Please answer if you know]

Answer 1

A. 0.502

B. 0.753

Step-by-step explanation:

A. The angular acceleration of the flywheel is equal to the change in angular velocity divided by the change in time, which is equal to (4.6-2.09)/5 which is 0.502.
B. Angular acceleration is equal to the net torque divided by the rotational inertia. If you rearrange the equation, torque is equal to angular acceleration times rotational inertia, which is 0.502 times 1.5 which is 0.753.

Answer 2

Assumption: the angular acceleration of this flywheel is constant.

Angular acceleration:
`0.502\; {\rm rad\cdot s^(-2)}`.

Net torque on this flywheel:
`0.753\; {\rm kg \cdot m^(2) \cdot s^(-2)` (or, equivalently,
`0.753\; {\rm N \cdot m}`.)

Step-by-step explanation:

Change in the angular velocity of this flywheel:

`\begin{aligned}\Delta \omega &= \omega_(1) - \omega_(0) \\ &= 4.6\; {\rm rad \cdot s^(-1)} - 2.09 \; {\rm rad \cdot s^(-1)} \\ &= 2.51\; {\rm rad \cdot s^(-1)}\end{aligned}`.

Divide this change in angular velocity
`\Delta \omega` with the duration
`t` of the acceleration to find the average angular acceleration of this flywheel:

`\begin{aligned}\alpha &= (\Delta \omega)/(t) \\ &= \frac{2.51\; {\rm rad \cdot s^(-1)}}{5\; {\rm s}} \\ &= 0.502\; {\rm rad \cdot s^(-2)}\end{aligned}`.

By Newton's Second Law for Rotation, the net torque on this flywheel would be the product of the angular acceleration and the moment of inertia of this flywheel,
`I`:

`\begin{aligned}\tau_{\text{net}} &= I\, \alpha \\ &= 1.5\; {\rm kg \cdot m^(2)} * 0.502\; {\rm s^(-2)} \\ &= 0.753\; {\rm kg \cdot m^(2) \cdot s^(-2)} \\ &= 0.753\; {\rm (kg \cdot m \cdot s^(-2)) \cdot m} \\ &= 0.753\; {\rm N \cdot m}\end{aligned}`.